In the attached sketch,
for the cyclic quadrilateral abcd, Q+S=P+R=180 degrees.
If "d" is outside the circle on the line ad, angle S decreases, while Q remains the same. If "d" is inside the circle on the same line, S increases while Q remains the same.
In both these cases, Q+S is no longer 180 degrees.
Neither is P+R.
All angles sum to 360 degrees, but opposite angles will no longer sum to 180 degrees. Hence, if opposite angles sum to 180 degrees, the quadrilateral is cyclic, as a circle circumference will pass through all 4 vertices.

Wow! I really started off on the wrong foot. I have good reason for being a bit confused and tired [medicines], but have not made mistakes like this for a while. Sorry about that. It won't happen again. I was somehow thinking "rectangle" instead of "quadrilateral". In future I'll stick to answering when not so tired out.

It does not matter if the quadrilateral includes the center or not. The proof is the same. With cyclic quadrilateral ABCD, the points A and C (or "arc AC") subtend angles at B and D. Each of those is one half of the angle subtended at the center [whether inside or out does not matter]. The angles at the center add to 360, so those at the circumference, B and D, add to 90. So then do A and C for several reasons, one of which is that the angles in any quadrilateral add to 360 in total.

EDIT: Just learning to upload an attachment. Please bear with me. This, if it works, is what I'm driving at:

This is my first posting in this forum, so I don't want to bend anyone out of shape. However, there have been some mistakes made. For example, an "inscribed" quadrilateral must have its four corners on the circle by definition. Therefore the center of the circle MUST be inside the quadrilateral.

You can show that if the four points of the quadrilateral all lie in the same 180-degree arc of the circle, that the center of the circle is not inside the quadrilateral.